Error: Trying to access a property that is not declared on an empty object

Question

Error: Trying to access a property that is not declared on an empty object

Using a fully patched Visual Studio 2013, I am integrating JQuery, JQueryUI, JSRender, and TypeScript into my project. However, I am encountering an error in the ts file:

Property 'fadeDiv' does not exist on type '{}'.

While I believe I have the necessary references for TypeScript, it seems like this issue stems from a d.ts problem.

Although there are no errors in JavaScript, Visual Studio keeps flagging 'fadeDiv' with red lines. The error message remains consistent:

/// <reference path="../scripts/typings/jquery/jquery.d.ts" />
/// <reference path="../scripts/typings/jqueryui/jqueryui.d.ts" />
/// <reference path="typings/jsrender/jsrender.d.ts" />

var SUCSS = {};

$(document).ready(function () {
   SUCSS.fadeDiv();
});

SUCSS.fadeDiv = function () {
var mFadeText: number;
$(function () {
    var mFade = "FadeText";
    //This part actually retrieves the info for the fadediv
    $.ajax({
        type: "POST",
        url: "/js/sucss/General.aspx/_FadeDivList",
        contentType: "application/json; charset=utf-8",
        dataType: "json",
        error: function (xhr, status, error) {
            // Show the error
        },
        success: function (msg) {
            mFadeText = msg.d.Fade;
            if (msg.d.FadeType == 0) {//FadeDivType = List
                var template = $.templates("#theTmpl");
                var htmlOutput = template.render(msg.d);
                $("[id$=lblFadeDiv]").html(htmlOutput);
            }
            else {//FadeDivType = String
                $("[id$=lblFadeDiv]").html(msg.d.FadeDivString);
            }
        },
        complete: function () {
            if (mFadeText == 0) {
                $("[id$=lblFadeDiv]").fadeIn('slow').delay(5000).fadeOut('slow');
            }
        }
    });
});

In TypeScript, to make 'SUCSS.fadeDiv' accessible externally, the following structure would be appropriate:

$(document).ready(function () {
    SUCSS.fadeDiv();
});
module SUCSS {
    export function fadeDiv () {};
};

By exporting the function using 'export', one can call 'SUCSS.fadeDiv' at page load with the syntax 'SUCSS.fadeDiv();'. This explanation may serve as a helpful guide.

typescript

Answer 1

Answer №1

` element, you have the option to set the `any` type for the specified object: To illustrate:

let variable: any = {};
variable.property = "value";

Answer 2

` element, you have the option to set the `any` type for the specified object: To illustrate:

let variable: any = {};
variable.property = "value";

Answer 3

Answer №2

To bypass strict type checking for a single field, use array notation:

data['propertyName']; //this will still work even if propertyName is not declared in data

Another approach is to (un)cast the variable for individual access:

(<any>data).propertyName;//access propertyName as if data has no specific type

The first method is more concise, while the second method is clearer about (un)casting types

You can also completely disable type checking for all fields of a variable:

let untypedVariable:any= <any>{}; //turn off type checking when declaring the variable
untypedVariable.propertyName = anyValue; //all fields in untypedVariable can be assigned and read without type checking

Note: Disabling type checking for all fields is riskier than just doing it for a single field, as all subsequent accesses on any field will be untyped.

Answer 4

To bypass strict type checking for a single field, use array notation:

data['propertyName']; //this will still work even if propertyName is not declared in data

Another approach is to (un)cast the variable for individual access:

(<any>data).propertyName;//access propertyName as if data has no specific type

The first method is more concise, while the second method is clearer about (un)casting types

You can also completely disable type checking for all fields of a variable:

let untypedVariable:any= <any>{}; //turn off type checking when declaring the variable
untypedVariable.propertyName = anyValue; //all fields in untypedVariable can be assigned and read without type checking

Note: Disabling type checking for all fields is riskier than just doing it for a single field, as all subsequent accesses on any field will be untyped.

Answer 5

Answer №3

const propName = data['propName'];

Answer 6

const propName = data['propName'];

Answer 7

Answer №4

When you input the code line below into TypeScript:

var SUCSS = {};

The type of SUCSS is automatically determined based on the assignment (it becomes an empty object type).

Later, when you try to add a property to this type, such as:

SUCSS.fadeDiv = //...

The compiler issues a warning indicating that there is no fadeDiv property in the SUCSS object (this kind of warning can help identify typos).

You have two options: either define the type of SUCSS explicitly (which does not allow assigning {}, since it wouldn't match the specified type):

var SUCSS : {fadeDiv: () => void;};

Or, assign the complete value initially and let TypeScript infer the types:

var SUCSS = {
    fadeDiv: function () {
        // Simplified version
        alert('Called my func');
    }
};

Answer 8

When you input the code line below into TypeScript:

var SUCSS = {};

The type of SUCSS is automatically determined based on the assignment (it becomes an empty object type).

Later, when you try to add a property to this type, such as:

SUCSS.fadeDiv = //...

The compiler issues a warning indicating that there is no fadeDiv property in the SUCSS object (this kind of warning can help identify typos).

You have two options: either define the type of SUCSS explicitly (which does not allow assigning {}, since it wouldn't match the specified type):

var SUCSS : {fadeDiv: () => void;};

Or, assign the complete value initially and let TypeScript infer the types:

var SUCSS = {
    fadeDiv: function () {
        // Simplified version
        alert('Called my func');
    }
};

Answer 9

Answer №5

Here is a recommendation for adjustment

declare const propertyName: any;

Answer 10

Here is a recommendation for adjustment

declare const propertyName: any;

Answer 11

Answer №6

You have the option to use the partial utility type in order to make all of the object's properties optional. Click here for more information

type MyObject = {
  name: string;
}

const myObj: Partial<MyObject> = {}
myObj.name = "John"

Answer 12

You have the option to use the partial utility type in order to make all of the object's properties optional. Click here for more information

type MyObject = {
  name: string;
}

const myObj: Partial<MyObject> = {}
myObj.name = "John"

Answer 13

Answer №7

updateParams(
        inputValues : {
            value1: any,
            value2: string
            value3: string          
        }){
          inputValues.value1 = inputValues.value1 + 'canUpdate';
          //inputValues.value4 = "Unchangeable";
          var updatedValues : any = inputValues;// losing the typescript on the new object of type any
          updatedValues.value4 =  'canUpdate';
          return updatedValues;
      }

Answer 14

updateParams(
        inputValues : {
            value1: any,
            value2: string
            value3: string          
        }){
          inputValues.value1 = inputValues.value1 + 'canUpdate';
          //inputValues.value4 = "Unchangeable";
          var updatedValues : any = inputValues;// losing the typescript on the new object of type any
          updatedValues.value4 =  'canUpdate';
          return updatedValues;
      }

Answer 15

Answer №8

var SUCSS = {}; implicitly sets the type of SUCSS as an object with no properties. To allow optional properties, you need to explicitly define its type.

type SUCESSType = {
    fadeDiv?: () => void;
};

const SUCSS: SUCESSType = {};

However, since the value of fadeDiv is defined immediately afterwards, there is no need for this property to be optional (which would require checking if it exists before calling it).

You can simply assign the function when creating the object.

const SUCSS = {
    fadeDiv = function () { /* function body */ }
};

Answer 16

var SUCSS = {}; implicitly sets the type of SUCSS as an object with no properties. To allow optional properties, you need to explicitly define its type.

type SUCESSType = {
    fadeDiv?: () => void;
};

const SUCSS: SUCESSType = {};

However, since the value of fadeDiv is defined immediately afterwards, there is no need for this property to be optional (which would require checking if it exists before calling it).

You can simply assign the function when creating the object.

const SUCSS = {
    fadeDiv = function () { /* function body */ }
};

Answer 17

Answer №9

Make sure to include var fadeDiv = ... at the beginning of your file, rather than simply writing fadeDiv = .... This will properly declare the variable.

You might be receiving the error "

Property 'fadeDiv' does not exist on type '{}'.

" for a line that you have not shared in your example. It appears that there is no reference to a fadeDiv property in the snippet provided.

Answer 18

Make sure to include var fadeDiv = ... at the beginning of your file, rather than simply writing fadeDiv = .... This will properly declare the variable.

You might be receiving the error "

Property 'fadeDiv' does not exist on type '{}'.

" for a line that you have not shared in your example. It appears that there is no reference to a fadeDiv property in the snippet provided.

Error: Trying to access a property that is not declared on an empty object

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

Answer №9

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